Let $$y = y(x)$$ be the solution curve of the differential equation $$\frac{dy}{dx} + \frac{2x^2+11x+13}{x^3+6x^2+11x+6}y = \frac{x+3}{x+1}$$, $$x > -1$$, which passes through the point (0, 1). Then $$y(1)$$ is equal to

We have the linear differential equation $$\frac{dy}{dx} + \frac{2x^2 + 11x + 13}{x^3 + 6x^2 + 11x + 6}\,y = \frac{x + 3}{x + 1}$$ with $$x > -1$$ and $$y(0) = 1$$.

We factor the denominator: $$x^3 + 6x^2 + 11x + 6 = (x+1)(x+2)(x+3)$$. Now we decompose $$\frac{2x^2 + 11x + 13}{(x+1)(x+2)(x+3)}$$ into partial fractions. Setting $$\frac{2x^2 + 11x + 13}{(x+1)(x+2)(x+3)} = \frac{A}{x+1} + \frac{B}{x+2} + \frac{C}{x+3}$$, we get $$2x^2 + 11x + 13 = A(x+2)(x+3) + B(x+1)(x+3) + C(x+1)(x+2)$$.

Substituting $$x = -1$$: $$2 - 11 + 13 = A(1)(2) \Rightarrow 4 = 2A \Rightarrow A = 2$$. Substituting $$x = -2$$: $$8 - 22 + 13 = B(-1)(1) \Rightarrow -1 = -B \Rightarrow B = 1$$. Substituting $$x = -3$$: $$18 - 33 + 13 = C(-2)(-1) \Rightarrow -2 = 2C \Rightarrow C = -1$$.

So the integrating factor is $$\mu = e^{\int P(x)\,dx}$$ where $$P(x) = \frac{2}{x+1} + \frac{1}{x+2} - \frac{1}{x+3}$$. Hence $$\int P\,dx = 2\ln|x+1| + \ln|x+2| - \ln|x+3| = \ln\!\left(\frac{(x+1)^2(x+2)}{x+3}\right)$$, and $$\mu = \frac{(x+1)^2(x+2)}{x+3}$$.

Multiplying the ODE by the integrating factor, we get $$\frac{d}{dx}\!\left[\frac{(x+1)^2(x+2)}{x+3}\,y\right] = \frac{x+3}{x+1} \cdot \frac{(x+1)^2(x+2)}{x+3} = (x+1)(x+2)$$. Integrating the right side: $$\int (x+1)(x+2)\,dx = \int (x^2 + 3x + 2)\,dx = \frac{x^3}{3} + \frac{3x^2}{2} + 2x + C$$.

Therefore $$\frac{(x+1)^2(x+2)}{x+3}\,y = \frac{x^3}{3} + \frac{3x^2}{2} + 2x + C$$. Using $$y(0) = 1$$: $$\frac{(1)(2)}{3} \cdot 1 = 0 + C$$, so $$C = \frac{2}{3}$$.

Now at $$x = 1$$: $$\frac{(2)^2(3)}{4}\,y(1) = \frac{1}{3} + \frac{3}{2} + 2 + \frac{2}{3}$$. The left side equals $$3\,y(1)$$. The right side equals $$\frac{1}{3} + \frac{2}{3} + \frac{3}{2} + 2 = 1 + \frac{3}{2} + 2 = \frac{9}{2}$$. Hence $$y(1) = \frac{9}{2} \cdot \frac{1}{3} = \frac{3}{2}$$.

Hence, the correct answer is Option B.